v = v0 + at
1D Kinematics Equations for Constant Acceleration
Physics 103: Lecture 5 Vectors - Motion in Two Dimensions
Today’s lecture will be on Vectors Two dimensions » projectile motion
! !
2/6/08
! 1
Physics 103, Spring 2008, U.Wisconsin
_
x = x 0 + vt
Δx = vt
x = x 0 + v 0 t + 12 at 2
Δx = v0t + 1/2 at2 (parabolic)
v = v 0 + at
Δv = at
v 2 " v 02 = 2a(x " x 0 ) 2/6/08
(linear)
(linear)
v2 = v02 + 2a Δx (independent of time)
2
Physics 103, Spring 2008, U.Wisconsin
! Lecture 3, Pre-Flight Q.3
Free-Fall
constant downward acceleration
g: acceleration due to gravity
same for all bodies: g=9.8 m/s2
ay = -g = -9.8 m/s2
up y
Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground? 1. Dennis' ball 2. Carmen's ball Dennis Carmen v0 3. Same Correct: v 2 = v02 -2gΔy v0
x
down
H
Summary of Free-Fall Equations y = y0 + v0yt - 1/2 gt2
vA
On the dotted line: Δy=0 ==> v2 = v02
vB
v = v0- g t
v y = v0y - gt
v = ±v0
t = 2 v0 / g
v y2 = v0y2 - 2gΔy
When Dennis’s ball returns
v = -v0
to dotted line its v = -v0 Same as Carmen’s 2/6/08
Physics 103, Spring 2008, U.Wisconsin
3
2/6/08
Free Fall Scenarios
A battleship simultaneously fires two shells at enemy ships from identical canons. If the shells follow the parabolic trajectories shown, which ship gets hit first?
Is the motion symmetrical? Then tup = t down Then v = -vo
1. Ship A.
Higher the shell flies, the longer it takes.
2. Ship B.
3. Both at the same time
The motion may not be symmetrical ⇒ Break the motion into various segments Are there symmetrical segments? Possibilities include » Upward and downward portions » symmetrical portion back to release point and non-symmetrical portion
What should the captain order if he wants to hit both ships at the same time?
A 2/6/08
Physics 103, Spring 2008, U.Wisconsin
4
Physics 103, Spring 2008, U.Wisconsin
5
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
B 6
Vector Algebra
Two Dimensions
Position can be anywhere in the plane Select an origin Draw two mutually perpendicular lines meeting at the origin Select +/- directions for horizontal (x) and vertical (y) axes Any position in the plane is given by two signed numbers
A vector points to this position
Analytical method Add the components separately to get the components of sum vector » Rx = R1x + R2x » Ry = R1y + R2y Scalar multiplication of vector » Can change magnitude and sign
R x 2+ R y 2 The angle of that vector is, θ = tan-1( Ry / Rx ) The square of its length is, R2=
R
Ry
Rx
2/6/08
Ry = R sin θ
Physics 103, Spring 2008, U.Wisconsin
7
2/6/08
R2
R1
Physics 103, Spring 2008, U.Wisconsin
ax = 0
x = x0 + v0xt + 1/2 ax t 2
y = y0 + v0yt + 1/2 ay t 2
vx = v0x + ax t
vy = v0y + ay t
x = x0 + v xt
y = y0 + v0yt - 1/2 gt2
vx2 = v0x2 + 2ax Δx
vy2 = v0y2 + 2ay Δy
vx = v0x
vy = v0y - gt
vy2 = v0y2 - 2g Δy
x and y motions are independent! They share a common time t
9
2/6/08
Physics 103, Spring 2008, U.Wisconsin
10
Projectile Motion: Maximum height reached Time taken for getting there
Question? Without air resistance, an object dropped from a plane flying at constant speed in a straight line will
Final velocity, v y = 0
1. Quickly lag behind the plane.
Height reached, hmax = y " y 0
2. Remain vertically under the plane.
2 Using kinematics equation, v y2 " v 0y = "2g( y " y 0 ) 2 v 0y 2g Time taken to reach this height, using v y " v 0y " gt,
3. Move ahead of the plane There is no acceleration along horizontal - object continues to travel at constant speed (same as that of the plane) along horizontal. Due to gravitational acceleration the object’s speed downwards increases. Physics 103, Spring 2008, U.Wisconsin
8
ay = -g = -9.8
Physics 103, Spring 2008, U.Wisconsin
x
Kinematics for Projectile Motion
x and y motions are independent! They share a common time t
2/6/08
D=R2-R1
D
Kinematics in Two Dimensions
2/6/08
R=R1+R2
y
Negation of vector (multiplying by -1) » Reverse signs of both components » Vector points in opposite direction
Rx = R cos θ
θ
Multiply all components by scalar – Components of sR are sRx and sRy
hmax =
v 0y g Depends only on the vertical component of the initial velocity t max =
11
2/6/08
!
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
12
Projectile Motion: Maximum Range
Projectile Motion
2v 0y (twice the time to top) g Range is maximum distance traveled along horizontal axis 2v 2v sin " 0 R = v 0x t = v 0x 0y = v 0 cos" 0 0 g g Total time of travel, t =
R=
v 02 sin2" 0 g
v 0 = 25m /s,g = 9.8m /s2 ," = 30 o
v 02 sin2" 0 , using trig. id. sin2" = 2sin " cos" g Depends on both magnitude and direction of initial velocity
!
R=
R=
25 2 sin(60) = 55m 9.8
Maximum range is for sin2" = 1, i.e., " = 45 o
! !
2/6/08
13
Physics 103, Spring 2008, U.Wisconsin
2/6/08
Shooting the Monkey...
Shooting the Monkey...
x = v0 t y = -1/ 2 g t2 , g = 9.8
y = y 0 - 1/ 2 g t2
At an angle, still aim at the monkey!
x = x0 y = - 1 / 2 g t2
No monkeys were harmed during the making of this slide
2/6/08
15
Physics 103, Spring 2008, U.Wisconsin
Dart hits the monkey!
y = vy0 t - 1 /2 g t2 g = 9.8
2/6/08
Physics 103, Spring 2008, U.Wisconsin
Shooting the Enemy Paratrooper... If a soldier wants to shoot down an enemy paratrooper descending at uniform speed, s, where should he aim? 1. Above the enemy 2. At the enemy 3. Below the enemy 4. Answer depends on the enemy’s position and vertical speed.
Projectile Motion
y = y 0 - st
= y0 + v0yt - 1/2 gt2
x
= x0 + v0t
y
v
= v0x
vy
R=
y = vy0 t + 1 / 2 g t2 Miss the enemy
Physics 103, Spring 2008, U.Wisconsin
16
Summary
= v0y - gt
vy 2
2/6/08
14
Physics 103, Spring 2008, U.Wisconsin
17
2/6/08
v 02 sin2" 0 g
Physics 103, Spring 2008, U.Wisconsin
!
Page ‹#›
= v0y2 - 2g Δy
18
Reference Frames: Relative Motion
Relative Motion
VCB=VCA+VAB
If an airplane flies in a jet stream, depending on the relative orientation of the airplane and the jet stream, the plane can go faster or slower than it normally would in the absence of the jet stream If a person rows a boat across a rapidly flowing river and tries to head directly for the shore, the boat moves diagonally relative to the shore Velocity is a vector - add velocities like vectors Sum the components » Vx = V1x + V2x
V1x = V1 cosθ
» Vy = V1y + V2y
Velocity of B relative to ground ( C ) : VCB
Velocity of A relative to ground ( C ) : VCA
V1y = V1 sinθ
Velocity of B relative to A : V AB 2/6/08
Physics 103, Spring 2008, U.Wisconsin
19
2/6/08
Lecture 5, Pre-Flight 3&4
20
Lecture 5, Pre-Flight 3&4
Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and Carly (C) swims downstream. Which swimmer wins the race? A) Ann B) Beth C) Carly
Physics 103, Spring 2008, U.Wisconsin
Beth will reach the shore first because the vertical component of her velocity is greater than that of the other swimmers. The key here is how fast the vector in the vertical direction is. "B" focuses all of its speed on the vertical vector, while the others divert some of their speed to the horizontal vectors.
correct
A B C
A B C
Time to get across = width of river/vertical component of velocity 2/6/08
Physics 103, Spring 2008, U.Wisconsin
21
2/6/08
A seagull flies through the air with a velocity of 9 m/s if there were no wind. However, it is making the same effort and flying in a headwind. If it takes the bird 20 minutes to travel 6 km as measured on the earth, what is the velocity of the wind?
A boat is drifting in a river which has a current of 1 mph. The boat is a half mile upstream of a rock when an observer on the boat sees a seagull overhead. The observer sees the gull flying continuously back and forth at constant speed (10 mph) between the boat and the rock. When the boat passes close by the rock, how far (what distance) has the gull flown?
2/6/08
1. 4 m/s 2. -4 m/s
correct
3. 13 m/s 4. -13 m/s
correct 5 miles 10 miles Not sufficient information to determine the distance
Physics 103, Spring 2008, U.Wisconsin
22
Lecture 5, Pre-Flight 5 and 6
More Relative Motion:
• • •
Physics 103, Spring 2008, U.Wisconsin
• Seagull’s velocity relative to the wind = 9 m/s • i. e., in the frame relative to the wind, wind velocity is zero • Seagull travels at 6000/1200 = 5 m/s relative to earth. Therefore, the wind velocity relative to earth is 5-9=-4 m/s. 23
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
24
Follow-up, Pre-Flight 5 and 6
Follow-up 2, Pre-Flight 5 and 6
If the seagull turns around and flies back how long will it take to return?
How are the round-trip times with and without wind related if the seagull always goes at 9 m/s?
1. More time than for flying out 2. Less time than for flying out
1. The round-trip time is the same with/without the wind 2. The round trip time is always larger with the wind
correct
3. The same amount of time
3. It is not possible to calculate this
• Seagull’s return velocity is: -4-9=-13 m/s. The speed is higher so it takes less time to return. Time taken for the return is given by 6000 m / 13 (m/s) = 461.5 s = 461.5/60 = 7.69 minutes 2/6/08
Physics 103, Spring 2008, U.Wisconsin
correct
• Time taken for the round trip with wind is: 27.69 minutes • Time taken for the round trip without wind is: 12000 m / 9 m/s = 1333 s = 22.2 minutes 25
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
26
Physics 103: Lecture 5 Vectors - Motion in Two Dimensions
Today’s lecture will be on Vectors Two dimensions » projectile motion
! !
2/6/08
! 1
Physics 103, Spring 2008, U.Wisconsin
_
x = x 0 + vt
Δx = vt
x = x 0 + v 0 t + 12 at 2
Δx = v0t + 1/2 at2 (parabolic)
v = v 0 + at
Δv = at
v 2 " v 02 = 2a(x " x 0 ) 2/6/08
(linear)
(linear)
v2 = v02 + 2a Δx (independent of time)
2
Physics 103, Spring 2008, U.Wisconsin
! Lecture 3, Pre-Flight Q.3
Free-Fall
constant downward acceleration
g: acceleration due to gravity
same for all bodies: g=9.8 m/s2
ay = -g = -9.8 m/s2
up y
Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground? 1. Dennis' ball 2. Carmen's ball Dennis Carmen v0 3. Same Correct: v 2 = v02 -2gΔy v0
x
down
H
Summary of Free-Fall Equations y = y0 + v0yt - 1/2 gt2
vA
On the dotted line: Δy=0 ==> v2 = v02
vB
v = v0- g t
v y = v0y - gt
v = ±v0
t = 2 v0 / g
v y2 = v0y2 - 2gΔy
When Dennis’s ball returns
v = -v0
to dotted line its v = -v0 Same as Carmen’s 2/6/08
Physics 103, Spring 2008, U.Wisconsin
3
2/6/08
Free Fall Scenarios
A battleship simultaneously fires two shells at enemy ships from identical canons. If the shells follow the parabolic trajectories shown, which ship gets hit first?
Is the motion symmetrical? Then tup = t down Then v = -vo
1. Ship A.
Higher the shell flies, the longer it takes.
2. Ship B.
3. Both at the same time
The motion may not be symmetrical ⇒ Break the motion into various segments Are there symmetrical segments? Possibilities include » Upward and downward portions » symmetrical portion back to release point and non-symmetrical portion
What should the captain order if he wants to hit both ships at the same time?
A 2/6/08
Physics 103, Spring 2008, U.Wisconsin
4
Physics 103, Spring 2008, U.Wisconsin
5
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
B 6
Vector Algebra
Two Dimensions
Position can be anywhere in the plane Select an origin Draw two mutually perpendicular lines meeting at the origin Select +/- directions for horizontal (x) and vertical (y) axes Any position in the plane is given by two signed numbers
A vector points to this position
Analytical method Add the components separately to get the components of sum vector » Rx = R1x + R2x » Ry = R1y + R2y Scalar multiplication of vector » Can change magnitude and sign
R x 2+ R y 2 The angle of that vector is, θ = tan-1( Ry / Rx ) The square of its length is, R2=
R
Ry
Rx
2/6/08
Ry = R sin θ
Physics 103, Spring 2008, U.Wisconsin
7
2/6/08
R2
R1
Physics 103, Spring 2008, U.Wisconsin
ax = 0
x = x0 + v0xt + 1/2 ax t 2
y = y0 + v0yt + 1/2 ay t 2
vx = v0x + ax t
vy = v0y + ay t
x = x0 + v xt
y = y0 + v0yt - 1/2 gt2
vx2 = v0x2 + 2ax Δx
vy2 = v0y2 + 2ay Δy
vx = v0x
vy = v0y - gt
vy2 = v0y2 - 2g Δy
x and y motions are independent! They share a common time t
9
2/6/08
Physics 103, Spring 2008, U.Wisconsin
10
Projectile Motion: Maximum height reached Time taken for getting there
Question? Without air resistance, an object dropped from a plane flying at constant speed in a straight line will
Final velocity, v y = 0
1. Quickly lag behind the plane.
Height reached, hmax = y " y 0
2. Remain vertically under the plane.
2 Using kinematics equation, v y2 " v 0y = "2g( y " y 0 ) 2 v 0y 2g Time taken to reach this height, using v y " v 0y " gt,
3. Move ahead of the plane There is no acceleration along horizontal - object continues to travel at constant speed (same as that of the plane) along horizontal. Due to gravitational acceleration the object’s speed downwards increases. Physics 103, Spring 2008, U.Wisconsin
8
ay = -g = -9.8
Physics 103, Spring 2008, U.Wisconsin
x
Kinematics for Projectile Motion
x and y motions are independent! They share a common time t
2/6/08
D=R2-R1
D
Kinematics in Two Dimensions
2/6/08
R=R1+R2
y
Negation of vector (multiplying by -1) » Reverse signs of both components » Vector points in opposite direction
Rx = R cos θ
θ
Multiply all components by scalar – Components of sR are sRx and sRy
hmax =
v 0y g Depends only on the vertical component of the initial velocity t max =
11
2/6/08
!
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
12
Projectile Motion: Maximum Range
Projectile Motion
2v 0y (twice the time to top) g Range is maximum distance traveled along horizontal axis 2v 2v sin " 0 R = v 0x t = v 0x 0y = v 0 cos" 0 0 g g Total time of travel, t =
R=
v 02 sin2" 0 g
v 0 = 25m /s,g = 9.8m /s2 ," = 30 o
v 02 sin2" 0 , using trig. id. sin2" = 2sin " cos" g Depends on both magnitude and direction of initial velocity
!
R=
R=
25 2 sin(60) = 55m 9.8
Maximum range is for sin2" = 1, i.e., " = 45 o
! !
2/6/08
13
Physics 103, Spring 2008, U.Wisconsin
2/6/08
Shooting the Monkey...
Shooting the Monkey...
x = v0 t y = -1/ 2 g t2 , g = 9.8
y = y 0 - 1/ 2 g t2
At an angle, still aim at the monkey!
x = x0 y = - 1 / 2 g t2
No monkeys were harmed during the making of this slide
2/6/08
15
Physics 103, Spring 2008, U.Wisconsin
Dart hits the monkey!
y = vy0 t - 1 /2 g t2 g = 9.8
2/6/08
Physics 103, Spring 2008, U.Wisconsin
Shooting the Enemy Paratrooper... If a soldier wants to shoot down an enemy paratrooper descending at uniform speed, s, where should he aim? 1. Above the enemy 2. At the enemy 3. Below the enemy 4. Answer depends on the enemy’s position and vertical speed.
Projectile Motion
y = y 0 - st
= y0 + v0yt - 1/2 gt2
x
= x0 + v0t
y
v
= v0x
vy
R=
y = vy0 t + 1 / 2 g t2 Miss the enemy
Physics 103, Spring 2008, U.Wisconsin
16
Summary
= v0y - gt
vy 2
2/6/08
14
Physics 103, Spring 2008, U.Wisconsin
17
2/6/08
v 02 sin2" 0 g
Physics 103, Spring 2008, U.Wisconsin
!
Page ‹#›
= v0y2 - 2g Δy
18
Reference Frames: Relative Motion
Relative Motion
VCB=VCA+VAB
If an airplane flies in a jet stream, depending on the relative orientation of the airplane and the jet stream, the plane can go faster or slower than it normally would in the absence of the jet stream If a person rows a boat across a rapidly flowing river and tries to head directly for the shore, the boat moves diagonally relative to the shore Velocity is a vector - add velocities like vectors Sum the components » Vx = V1x + V2x
V1x = V1 cosθ
» Vy = V1y + V2y
Velocity of B relative to ground ( C ) : VCB
Velocity of A relative to ground ( C ) : VCA
V1y = V1 sinθ
Velocity of B relative to A : V AB 2/6/08
Physics 103, Spring 2008, U.Wisconsin
19
2/6/08
Lecture 5, Pre-Flight 3&4
20
Lecture 5, Pre-Flight 3&4
Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and Carly (C) swims downstream. Which swimmer wins the race? A) Ann B) Beth C) Carly
Physics 103, Spring 2008, U.Wisconsin
Beth will reach the shore first because the vertical component of her velocity is greater than that of the other swimmers. The key here is how fast the vector in the vertical direction is. "B" focuses all of its speed on the vertical vector, while the others divert some of their speed to the horizontal vectors.
correct
A B C
A B C
Time to get across = width of river/vertical component of velocity 2/6/08
Physics 103, Spring 2008, U.Wisconsin
21
2/6/08
A seagull flies through the air with a velocity of 9 m/s if there were no wind. However, it is making the same effort and flying in a headwind. If it takes the bird 20 minutes to travel 6 km as measured on the earth, what is the velocity of the wind?
A boat is drifting in a river which has a current of 1 mph. The boat is a half mile upstream of a rock when an observer on the boat sees a seagull overhead. The observer sees the gull flying continuously back and forth at constant speed (10 mph) between the boat and the rock. When the boat passes close by the rock, how far (what distance) has the gull flown?
2/6/08
1. 4 m/s 2. -4 m/s
correct
3. 13 m/s 4. -13 m/s
correct 5 miles 10 miles Not sufficient information to determine the distance
Physics 103, Spring 2008, U.Wisconsin
22
Lecture 5, Pre-Flight 5 and 6
More Relative Motion:
• • •
Physics 103, Spring 2008, U.Wisconsin
• Seagull’s velocity relative to the wind = 9 m/s • i. e., in the frame relative to the wind, wind velocity is zero • Seagull travels at 6000/1200 = 5 m/s relative to earth. Therefore, the wind velocity relative to earth is 5-9=-4 m/s. 23
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
24
Follow-up, Pre-Flight 5 and 6
Follow-up 2, Pre-Flight 5 and 6
If the seagull turns around and flies back how long will it take to return?
How are the round-trip times with and without wind related if the seagull always goes at 9 m/s?
1. More time than for flying out 2. Less time than for flying out
1. The round-trip time is the same with/without the wind 2. The round trip time is always larger with the wind
correct
3. The same amount of time
3. It is not possible to calculate this
• Seagull’s return velocity is: -4-9=-13 m/s. The speed is higher so it takes less time to return. Time taken for the return is given by 6000 m / 13 (m/s) = 461.5 s = 461.5/60 = 7.69 minutes 2/6/08
Physics 103, Spring 2008, U.Wisconsin
correct
• Time taken for the round trip with wind is: 27.69 minutes • Time taken for the round trip without wind is: 12000 m / 9 m/s = 1333 s = 22.2 minutes 25
2/6/08
Page ‹#›
Physics 103, Spring 2008, U.Wisconsin
26
